Timeline for Showing that the inverse of a function is approximately equivalent to $\frac{1}{n^{1/\alpha}}$
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Nov 19, 2020 at 15:26 | comment | added | Andrew Larkin | Pietro Majer: This is the formula you're talking about right? en.wikipedia.org/wiki/Lagrange_inversion_theorem Do you take $f(x)$ as already defined, and $a = 0$? | |
| Nov 19, 2020 at 13:18 | comment | added | Andrew Larkin | 1) Ah yes, I think I had a brain fart and was confusing the fact that $g(x) < x$ with meaning that it was decreasing, obviously if you increase $x$ then the value of $g(x)$ will also increase. 2) Thank you, I'll try to derive that myself. | |
| Nov 19, 2020 at 13:16 | vote | accept | Andrew Larkin | ||
| Nov 19, 2020 at 12:08 | comment | added | Pietro Majer | 1) it's a general elementary fact, think to $\exp$ and $\log$ or $\tan$ and $\arctan$. 2) it's the Lagrange inversion formula, for the inverse of a power series. | |
| Nov 19, 2020 at 10:35 | comment | added | Andrew Larkin | Hi, thank you so much for your answer. I just have a couple of questions: 1) Since $f$ is an increasing function, how can $g$ also be increasing? 2) How did you obtain that explicit form of $g$? I imagine it's through Taylor expansion, right? Is there a specific form for the inverse of a function? | |
| Nov 19, 2020 at 2:02 | history | edited | Pietro Majer | CC BY-SA 4.0 |
added 149 characters in body
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| Nov 19, 2020 at 1:28 | history | answered | Pietro Majer | CC BY-SA 4.0 |